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12:01 AM
I've never heard of that and there appears to be no work on this except for this one person
 
There's a famous paper by Fadeev and Kulish
 
I strongly advise to learn standard QFT before delving into non-mainstream attempts to circumvent its weirdness
 
This asymptotic states thing is still bugging me after a year and I don't know what to do
 
Then do nothing
 
This seems to be an issue noone understands except very few people but everyone acts like they understand it
 
12:16 AM
Because they have full confidence in their understanding of the standard approach
 
The standard approach is just an incorrect placeholder just to get onto scattering quicker in qft courses
 
Experiments are where the proofs are.
 
Everything else in QFT mostly logically flows except this
I think it's done properly in Weinberg
I'm going to look there
 
12:40 AM
@ACuriousMind so given that I fix my conventions, I should be able to travel between the irreps in terms of single tensor products of irreps of sl(2, C) with conjugate irreps of sl(2, C) and the irreps in terms of single tensor products of irreps of $su(2)_\mathbb{C}$?
And then this should really give the appropriate concrete matrices
 
The crux of the issue is that we can't remove the interaction
Therefore another argument needs to be in place
 
 
3 hours later…
3:22 AM
@DIRAC1930 L&L's argument is crystal clear, please show me how Weinberg's comments on this are better than what they say
 
4:21 AM
hall says that The Lie algebra $g$ of a matrix Lie group $G$ is a real Lie algebra. I feel like I am misunderstanding this statement. Isn't $gl(n;\mathbb{C})$ a complex Lie algebra and the Lie algebra of the general linear group, which is a matrix Lie group?
 
4:57 AM
@Relativisticcucumber He is probably assuming every lie group, unless otherwise specified, is real
 
5:28 AM
@Relativisticcucumber what page is this on?
 
6:07 AM
Is it just me? Or is everyone still seeing the PSE in read-only mode?
 
@123 Sigh, you should not take what the other guy said. He has no idea what it is you are talking about. And no, you cannot have person 1 apply 10N on person 2 while person 2 apply 15N force on person 1. That is a violation of conservation of momentum. They must apply the same force on each other. This is also why K&K said that while person 2 is stronger, it is the weaker person 1 that limits how much force person 2 can apply. Any more, and person 1 will fly.
 
6:24 AM
@SillyGoose page 54 at the bottom
its prop 2.39
 
6:58 AM
@Relativisticcucumber for posterity, the statement as stated is indeed true. the import is that Hall then introduces Ado's theorem for real Lie algebras to then simply prove Lie's third theorem. A possible confusion is that a matrix Lie algebra can have complex-entried elements. Recall though that the field of scalars for the underlying vector space structure has nothing to do with from where the entries of the matrix come from.
 
@bolbteppa How do I put L&L's arguments into mathematical form i.e. I) Show that $\psi_p \sim e^{\imath p x}$ is a solution to the full interacting $\hat{H}$ and II) Show that the time dependence of $\psi_p$ is determined by the single-particle wave equation?
I think I should understand this archive.org/details/…
THE relativistic dynamics of a free particle is fully determined by the behavior of its state vector under transformations $\mathbf{g}$ of the quantum mechanical Poincaré group $\mathcal{D}^{\uparrow.}_+$. Knowledge of the explicit form of the operator $U(\mathbf{g})$ acting on the particle state vector is equivalent to the solution of the equation of motion for a free particle. In particular, after constructing $U(\mathbf{g})$ the transition to local fields becomes obvious.
 
7:24 AM
@naturallyInconsistent this is incorrect. the total net force applied by person 1 on 2 must be equal and opposite to the one applied by 2 on 1. if u read my msg, my example had two bodies applying two forces on each other at two different points of applications
 
@SillyGoose hm okay i think this makes sense.
thanks
 
these two forces are allowed to be different
 
7:49 AM
@RyderRude I am aware. You are the one who misinterpreted his question about N3L into something completely different. Stop replying.
 
where does the gl come into play here? i thought this map should be $\text{ad}: \mathfrak{g} \rightarrow \mathfrak{g}$
i guess im not sure how to interpret this gl also
 
8:17 AM
the adjoint map acts on a Lie algebra vector
So $\mathrm{ad}_{(-)}X : \mathfrak{g} \to \mathfrak{g}$, but if you don't specify a vector, it is only to a linear function on the algebra
 
@DIRAC1930 It's not a solution of the full interacting $H$, at $t = \pm \infty$ the interacting $H$ reduces to the free $H$ or there is nothing that can be measured which is the point of their whole argument, only free things can be measured
 
8:32 AM
@naturallyInconsistent the original question was without context of the rope. i interpreted it how it could be interpreted
 
No, it was clear from the first line that it was about N3L. You simply misread it, as is very common of you.
 
i knew it was about n3l, as my replies are also about n3l
it's just two action forces and two reaction forces, which is the only way it cud b interpreted as written
 
No, it is more sensible to interpret it as being inconsistent, because people who are confused about N3L are much more likely to be making mistakes in that vein.
 
that is subjective. i initially interpreted it as inconsistent..but it is more natural to interpret it as both people hitting each other with different forces
but anyway, it is subjective
 
I think he is moreso saying that a measurement of the particle over a time interval less than infinity would lead to the creation of electron-positron pairs rendering the statement that a specific particle you are measuring had momentum $p$ meaningless.

It is still a big leap to go from this to $\hat{H}$ at $t=\pm \infty$ reduces to the free $H$. We can still use particles obeying the full $\hat{H}$ to measure the momentum of particle $A$ by just using very low energy particles of the full $H$ to collide with particle $A$ meaning no electron positron pairs will be produced
But then maybe this is equivalent to switching interactions off
If you look at equation 117 here archive.org/details/…
He seems to construct that the one particle states obey $\mathbf{J} \cdot \mathbf{p} - J p^0$
 
8:44 AM
You are conflating things, and they literally say: "The only observable quantities are the properties (momenta, polarisations) of free particles: the initial particles which come into interaction, and the final particles which result from the process "
 
@bolbteppa But how do I show this mathematically
 
They showed it on the previous pages
 
That would be like trying to do all of QM just from the Heisenberg uncertainty principle
 
also, i feel like im not understanding the last line of this proof. this is to show that $XY - YX \in \mathfrak{g}$. i see the derivative stuff in the middle, and i see that $e^{tX}Ye^{-tX} \in \mathfrak{g}$, but im not getting why we can conclude that $XY - YX \in \mathfrak{g}$ from this ?
 
8:47 AM
The book I showed you shows I think that one particle states obey the single particle massless Dirac equation (equation 117) archive.org/details/…
 
That's basically how they do all of QM, from Heisenberg's uncertainty principle, even in the NRQM book
You're just cherry picking random things now, this book is setting up the Wigner method
 
See the introduction the chapter archive.org/details/…
It tries to set up equations of motion for the one particle states to show that they are free particles
 
Yes it's just setting up the Wigner method
 
@RyderRude You just cannot be wrong. It just has to be subjective. It just cannot become objective the moment he brought up K&K and the tug of war.
 
@bolbteppa Is this what I need?
 
8:50 AM
No the Wigner method is not going to let you bypass any of this
 
But it shows that the equations of motion for the one particle state are the free relativistic wave equations
i.e. Dirac equation
 
It's very simple, in come free particles, and out go free particles, which is exactly what we do in every QFT problem, this explains why that is the case, the only difference is every other book treats it as some problem arising from our ignorance and inability to solve complex equations, this book is telling you it's fundamental
 
@Relativisticcucumber This is the kind of situation whereby I kinda feel like I need to see the whole proof, i.e. points 1 2 and 3, before I am confident in posting a reply; I am really trying to ask you to improve how you take screenshots so as to ask better questions.
 
@naturallyInconsistent at that point, it did become objective..but i did not misinterpret what was initially written.
 
8:54 AM
@naturallyInconsistent i thought 1 and 2 are ind of 3, but i attached in any case
 
or rather, my interpretation was natural
 
@RyderRude Yet you did not go back to point out that your example clearly did not fit with what K&K was trying to say. And you still think you did not misinterpret him.
@ACuriousMind please act on this.
 
@bolbteppa This doesn't answer the question about why exactly a particle infront of me obeys the free rel wave equations
 
@naturallyInconsistent i thought it was clear that they were different situations
 
All it does is give an argument about why transition amplitudes between free particles are intuitively the right thing to consider
 
8:56 AM
@naturallyInconsistent otherwise, @123 wouldve asked me about the contradiction
 
@RyderRude stop replying nor tagging me.
 
Obviously, it tells you in general how every system of particles that can be measured is going to behave, then you study each system separately and study the dynamics of each system
 
@Relativisticcucumber Yes, it does seem like they are independent, but note that the proof of 3 referenced 1 and 2, so it actually is important to state them together.
 
i see
my bad
 
In order to do that, you consider representations of the Poincare group and study properties of those representations
Which is what that chapter is doing
 
9:00 AM
You're interpreting it all in hindsight. You could for example come to the conclusion that free particles can't exist because you can't switch interactions off. That logically follows, but because you know how calculations work in QFT, that seems like a silly thing to suggest even though it logically follows
 
@DIRAC1930 i think the claim is simply that asymptotic states approximately evolve according to the renormalised free hamiltonian..this is only approximate
i think the same situation exists in non rel QM too
far from the potential, the eigenstates approximately look like plane waves
 
If you couldn't switch interactions off, yes that would be true and the theory would completely collapse, that is the claim, nothing would be measurable, but obviously things are measurable and fields are assumed to die off as we go to infinity
 
@bolbteppa Ah thanks. So what you are saying is that L&L gives a reason why interactions can be switched off
 
@Relativisticcucumber I think I kinda made sense of what it is claiming now. First, you need to be a closed set. You need the closure because then the limit taking will not take you outside of the set. Then with point 1, you can show that if $Y\in\mathfrak g$, then $-Y\in\mathfrak g$, and then with point 2, the subtraction in the numerator is also in $\mathfrak g$, and then again with point 1, the division by scalar $h$ is also in $\mathfrak g$. Finally, the limit taking is allowed by the closed set.
 
Roughly what they are doing is constructing one particle states as irreducible representations of the Poincare group, showing that the states depend on momenta and a representation of the Little group which characterizes spin/helicity, and they normalize those states, then they magically re-write states in terms of creation operators acting on the vacuum, then use this normalization to show the operators can satisfy either bose/fermi commutation relations, and fields are just linear combinations
 
9:07 AM
@bolbteppa Where are they doing this in L&L?
 
No, they are assuming interactions can be switched off because free theories are assumed to always exist
@DIRAC1930 In the other book you linked to, which is similar to what Weinberg does though it was written like 20-30 years before
 
In equation 117, are those $\mathbf{J},\mathbf{p},p^0$ operators i.e. $p^0 = \hat{H}$?
 
They are eigenvalues of the state
 
So is this an equation of motion or not?
 
Not sure what he's saying there
 
9:17 AM
it those were operators, you would have the massless Dirac equation I think
i.e. the equation of motion for those states would be the single particle massless Dirac equation
 
@JohnRennie the site looks fine to me.
 
@user85795 It seems as though only some people are being affected.
 
🤔 hmmmm
 
@naturallyInconsistent bleb okay i see
wait
sorry actually im still not seeing how we know that $\mathfrak{g}$ is closed. @naturallyInconsistent
 
9:34 AM
@Relativisticcucumber I also have difficulty seeing that, but it kinda makes sense. What is happening is that we just, via points 1 and 2, showed that $\mathfrak g$ is a vector space. As in, if you have two different "vectors" $X$ and $Y$ both in $\mathfrak g$, then any linear combination of both of them is still in $\mathfrak g$. IIRC, such a vector space is topologically closed. I don't remember if it inherits it from $\mathbb R^n$, but it is.
 
@SillyGoose yes
@Relativisticcucumber a finite-dimensional vector space is in particular closed, and points 1 and 2 show that $\mathfrak{g}$ is a vector space
 
@ACuriousMind oh yay I'm correct!
 
9:54 AM
@ACuriousMind i thought that closure of a vector space is different from topologically closed
@SillyGoose we are at this issue of closures again XD
 
@Relativisticcucumber well, I really meant complete if you want to avoid the word "closed"
 
oh no a new term
 
but since you're dealing with matrix algebras, $\mathfrak{g}$ is a (topologically) closed subset of some n-by-n matrix algebra, being closed (as a subset in that matrix algebra) means being complete, and finite-dimensional sub vector spaces are always closed subspaces
 
is this different from cauchy completeness?
 
yes, I mean Cauchy completeness
the chain of reasoning here is: $\mathfrak{g}$ is a finite-dimensional subvectorspace => $\mathfrak{g}$ is closed => $\mathfrak{g}$ is (Cauchy) complete => that limit in the proof must lie inside $\mathfrak{g}$
 
10:00 AM
@ACuriousMind hm so you mean any finite vector space is complete?
 
@Relativisticcucumber finite*-dimensional*, but yes
 
@ACuriousMind hm and is this a result of the vector space being closed, or it's independent?
 
@Relativisticcucumber depends: you could state it without closedness: all finite-dimensional vector spaces of dimension $n$ carry the topology of $\mathbb{R}^n$ (any linear isomorphism to $\mathbb{R}^n$ is also continuous) and $\mathbb{R}^n$ is complete, so all finite-dimensional vector spaces are complete
but the text you're reading clearly has a different argument in mind, which is 1. finite-dimensional sub vector spaces of some other vector spaces are closed 2. closed subsets of complete spaces are complete 3. the space of n-by-n-matrices is complete
 
@ACuriousMind does the fact that the space of $n \times n$ matrices is complete stem from the fact that the reals and complex numbers are complete?
 
@Relativisticcucumber yes - in the end all that's happening here is that $\mathbb{R}^n$ is complete because $\mathbb{R}$ is
 
10:10 AM
@ACuriousMind oh wait is that what u mean with the first message
ok i think i see
@ACuriousMind but what connects them to $\mathbb{R}^n$? what if i have $M_n(\mathbb{Q})$ for finite $n$?
 
@Relativisticcucumber that's a $\mathbb{Q}^{n\times n}$
but we're talking about real vector spaces here - point 1 in your proof shows that $\mathfrak{g}$ is a real vector space
 
kms you are right
ok great
always worrying about nonproblems
 
Note that it is always a headache if you work with spaces like $\mathbb Q$ because it is non-complete. Things like $\sqrt2$ would be holes in those spaces. It would be very annoying.
 
@naturallyInconsistent bleb $\sqrt{2}$ is my favorite number
@naturallyInconsistent i fear im becoming to focused on weird mathematical anomalies that don't matter for physics. i must resist.
 
@Relativisticcucumber Then dont consider spaces that exclude them!
 
10:23 AM
@naturallyInconsistent good point
 
@Relativisticcucumber Yes, you must.
 
@bolbteppa In your opinion, is the argument in that book enough to show that one particle states by definition are free particles that obey the free Hamiltonian? And then the L&L arguments to show that these free particles are the only measurable things hence inferring that these correspond to asymptotic states with definite momentum
 
the username on PSE can be changed right?
 
@ekardnam_ yes, it's under Settings->Edit Profile
 
10:34 AM
is it valid that the elements of the algebra here are matrices of matrices?
i dont know how to make sense of this
 
@ACuriousMind wait but it changed on PSE but not here
 
@Relativisticcucumber this is standard "block matrix" notation
Feb 7 at 12:29, by ACuriousMind
@Relativisticcucumber it's caching! but there's a mod button to manually refresh the caches, reload the page and you should see your chat avatar now also in all its wobbly glory
 
@ACuriousMind i have seen this before but an algebra can be like this?
what is the role of the symplectic groups in physics?
 
@Relativisticcucumber what does it have to do with this being an algebra? It's a 2n-by-2n matrix, and you're writing it as a 2-by-2 block of n-by-n matrices
 
@Relativisticcucumber in classical mechanics its important
 
10:37 AM
@ACuriousMind i guess im wondering what to make of the fact that we can write it this way because i feel like there must be some information that this gives but i am not seeing it
 
basically canonical transformations are the ones that preserve the symplectic structure of the phase space
 
@Relativisticcucumber are you trying to understand the point of a definition again before you've read the rest of the chapter where you'll see it being used ;P
 
no i finished the chapter
 
@ACuriousMind oh its just cache okay, it changed now yay
 
i tried ur method of saving all of my q's til the end and this is one of the remaining questions of ch 2 hall
 
10:38 AM
@Relativisticcucumber and didn't it use this form of the matrices somewhere?
 
@DIRAC1930 Let's go with a yes for now
 
not to my knowledge
 
at the very least this presentation is something you could use to determine the dimension of the algebra
 
That book should seriously start off by considering more simple cases e.g. like L&L
 
10:54 AM
@DIRAC1930 which book?
also L&L is Landau?
 
L&L is Landau & Lifshitz yes
although many of the books were written by other authors as well
 
@DIRAC1930 oh i never saw this book before
 
I don't recommend it
 
then why dont you study on a different book?
weinberg does some stuff on that kinda stuff too
 
Some information you can't find anywhere for reasons I don't know why
 
10:59 AM
One of the few books which studies Veneziano stuff
 
Yes Weinberg does that it's just his notation takes too long to decipher
 
@lucabtz that's because DIRAC1930 is digging through every obscure QFT resource they can find in the hopes that there will be one that magically resolves all the unanswered questions in QFT :P
 
@bolbteppa i wouldnt say the veneziano amplitude is a must for a begginer in QFT
@ACuriousMind i see
aaahhh finding algebraic geometry concepts again studying
 
11:26 AM
@ACuriousMind Who knows
I'm pretty sure every GR problem is solved somewhere in MTW
 
11:36 AM
I feel like the 1938/1943 Heisenberg paper would have all the answers but I can't find an English translation
 
@DIRAC1930 in the time you've spent chasing obscure old papers you could've read Reed & Simon's scattering theory by now
 
For now I'm just satisfied with the L&L arguments alongside the arguments in that other Russian book
 
obviously you're not satisfied since you keep talking about the same things :P
 
11:51 AM
I think that modern books are far better to learn particle physics
Old stuff uses outdated language
And in many cases they try to give a meaning to something that doesn't have one :P
something=quantum stuff of course
 
I find the newer texts just hide everything under the rug
 
I have the opposite impression, actually :P
Of course choosing the right book
 
Plus I don't want to hear anything about anything other than the photon, electron, positron, neutron and proton lol
 
Book which doesn't exist
 
that's essentially saying you think QCD is false, which puts you firmly outside the mainstream
 
11:57 AM
@DIRAC1930 what do you mean? Neutron and proton are non-elementary
You can understand that even in QED grounds
 
I'm not denying QCD its just I'm not interested in it
 
then why on earth do you want to learn QFT :P
 
I have no idea anymore lol
 
the main application of relativistic QFT - and the one almost all texts on it pursue - is to construct the Standard Model of particle physics
 
Probably why I like all the pre-standard model textbooks then :p
 
12:01 PM
Yeah SM is not fun
I mean, it is before you quantize classical fields
(Yeah, my fun stops after writing the lagrangian :P)
 
I also want to focus mostly on photon, electron, positron, neutron and proton. I also want to focus on QED and only a little bit into QCD. These are good enough reasons to want to study QFT. But it does not mean to learn outdated QFT. Good treatments of modern QFT exists.
 
In all seriousness, after you see how emergent fields are quantized in cond-mat (e.g. phonons) it kind've makes you think that QFT is not fundamental at all since and is just a framework where any classical field can be quantized. However, the ones that follow from space-time e.g. spin-0 spin-1/2 spin-1 just feel more fundamental in a sense
However the way most cond-mat physicists do QFT ruins the whole subject
Sorry ignore my spin-0 spin-1/2 spin 1 comment
 
If anything cond-mat does QFT in a way that is more beginner friendly. If you think it ruins the whole subject, that is on you.
 
L&L 9 does it properly
 
@naturallyInconsistent is it though?
 
12:17 PM
@SirCrackpot They at least attempt to connect the QM stuff to QFT. Way more sensible than just throwing students into the difficulties of relativistic QFT and expect them to swim.
 
I learned relativistic QFT first. Learning that beginner friendly version in Many Body Theory made it more difficult to me :P
 
@naturallyInconsistent I agree L&L 3 and L&L 4 do QFT best in my opinion
 
Not the topic per se, rather the way everything is constructed seems more tidy in rel QFT to me
 
Many body people are way better at calculating physical things
 
@DIRAC1930 Yes, yes, the parts of L&L not having much to do with Landau seems to be better than the original set; and people who had not yet understood enough of the Standard Model could possibly write a better textbook on QFT than the people who had finally figured out the SM and what that entailed about QFT.
 
12:21 PM
@DIRAC1930 why do we have to immediately go to assertions of who's "better"? Certainly both the cond-mat expert computing relevant quantities for condensed matter systems and the hep expert computing relevant quantities for particle colliders are "good at calculating physical things"
 
@SirCrackpot Sure, that is because HEP ppl had run into so many headaches trying to construct QFT properly that they kinda had to teach the next generations how to avoid those headaches. That does not mean that tidy constructions are beginner friendly. Measure theory can be tidily constructed. That does not mean it is beginner friendly to teach basic calculus that way.
 
coming from the hep-th world i prefer the hep approach to QFT, however I found interesting the many body approach too. it is always nice to get different views over one theory
@DIRAC1930 i dont think this sentence makes any sense tbh
 
Hep-th approach and cond-mat approach are exactly the same the more you look into it
You can still get field operators that are linear combinations of creation and anihilation operators I think
I don't know much about this however
 
@DIRAC1930 in all my cond-mat courses people preferred using the non-local creation/annihilation operators to define stuff rather than the local field operators
 
I don't know what that means
 
12:30 PM
Is there any reason why determinants don't usually come up as invariants? E.g. why don't we ever see the determinant of the Ricci tensor (I mean the one with raised index $R^\mu{}_\nu$)
 
@SirCrackpot determinants are not invariants, they transform as densities
 
@DIRAC1930 during my cond-mat courses people preferred using creation/annihilation operators over field operators, for example to define correlators and such
@ACuriousMind you could get an invariant by multiplying by a suitable power of the determinant of the metric though (?)
 
it's possible
this sounds more like a question for @Slereah
@lucabtz I suspect that's because the non-relativistic and finite-volume theories of cond-mat can use those operators with impunity
 
There's a theory where the action is the determinant of the Riemann tensor
Bit of a weird theory though
 
really we'd love to phrase QFT in purely in terms of the free operators (and hence also c/a operators), but once again that's essentially forbidden by Haag's theorem
 
12:38 PM
@ACuriousMind yeah
@ACuriousMind i should really read more on that. I think the Wightman book mentioned something about Haag's theorem?
 
the expressions that connect the free and interacting formalisms in those cond-mat theories will generically diverge if you try their analogues in relativistic infinite-volume QFT (just like the operator that connects the free and interacting vacuum in the interaction picture diverges)
 
@ACuriousMind oh i see
 
@lucabtz chapter 4-5 is literally named "Haag's theorem and its generalizations" ;)
 
@ACuriousMind yeah i kinda remembered that when i saw the book at the library
might start reading that after i hand in the thesis
which is going to be on 8th of March
 
12:52 PM
@lucabtz This is because the electron field operators go like $\hat{\Psi}^\dagger = \sum_p \hat{a}^\dagger_p \psi^*_p$ therefore the free propagator goes like $\sum_{p p'} \langle 0 | \hat{a}^\dagger_p \hat{a}_{p'}|0\rangle$ or something I think
 
@DIRAC1930 i mean i know the reason, its not like i was confused by it or something. its just a difference between hep and cond-mat. hep people would rather work with field operators
 
Yes I noticed that too
 
you also miss positrons in your field
 
Positrons can be excluded from the non-rel field operator
 
and the sum in the propagator is wrong, what you get using the c/a operators should be the momentum space one particle green's function, it does depend on the momentum
@DIRAC1930 oh i though we were doing high energy
 
12:56 PM
Oh my comment was a reply to your comment "during my cond-mat courses people preferred using creation/annihilation operators over field operators, for example to define correlators and such"
 
@ACuriousMind wait, that's if you take the determinant of a $(2,0)$ tensor
Just like the trace wouldn't be invariant if you naively computed it like $R_{\mu\mu}$
 
@ACuriousMind If you look at L&L 9, it is exactly the same as L&L 4 with the entire framework essentially copied and pasted from L&L 4
I wish there existed a scientific field that was in the middle of hep-th and cond-mat-th
 
its funny that pants decompositions of puntured riemann surfaces essentially answer the meme question "how would a punctured Riemann surface wear pants?"
@DIRAC1930 you are making yourself to many questions too early in my opinion
 
What do you mean?
 
you should go through this even if it isnt so clear rn
go through and especially study renormalization
 
1:11 PM
@lucabtz Go through what?
 
this asymptotic state mess
just accept what you dont understand and continue
stuff may become more clear as you continue studying
i dont think it will help to get stuck on this when you still havent done some more interesting things such as renormalization
 
I've seen renormalization
Plus I find the asymptotic states thing interesting
 
oh my bad i thought you were at the begginings in qft
 
I'll take that as an insult :p
 
@DIRAC1930 and yet you apparently refuse to study formal solutions to it such as Haag-Ruelle theory, why?
 
1:15 PM
Because it's longer than a page
 
@DIRAC1930 its just because you were making a lot of questions on asymptotic states and representation of Poincaré group which are usually what you do in a first course in QFT
@DIRAC1930 it seems like you want a "longer than a page" explanation for them though
otherwise just read any standard book where the LSZ formula proof is one page long
 
@lucabtz People do the stuff in that book as a first course?
 
@DIRAC1930 i mean single particle states of relativistic QFT were done in my QFT I
 
u shud also read about non rel scattering.
not particle scattering
but more like colliding particles on a fixed target
 
and in general the representation theory of the Poincaré group was done in QFT I
it was done the "physicist way", but good enough for a person with some maths basics to understand it
@RyderRude also this I would assume you did, at least to some degree, in the quantum mechanics course no?
 
1:22 PM
I have no maths basics
Everyone on here is leaps ahead of me in maths knowledge
 
@DIRAC1930 maybe we found the part you need to study more then
 
@lucabtz yes
read it from Shankar chapter 19
basically, the energy eigenstates of a decaying potential look like plane waves at infinity in non rel scattering too
this shud also be clear from the WKB approximation
 
@RyderRude my beloved WKB approximation
 
@lucabtz I think the issue is that most people blaze through QFT to get to ST without even understanding the basics properly so simple questions seem trivial but they are actually incredibly subtle.
 
i remember when i first read schrodinger eqn, i thought its exact solution was the WKB solution
but later, the book mentioned the WKB thing and said it's only an approximation
 
1:29 PM
@DIRAC1930 most people dont do ST, its a minority in the whole physics community really
 
@Slereah Riemann or Ricci?
 
@RyderRude there exists a so-called "exact WKB method". It involves considering the whole WKB series, which is only asymptotic and then using Borel resummations
 
yeah.. but i was only considering the first term when i had the idea
 
@RyderRude The difference here though is that the potential dies off at infinity i.e. there is physically no interaction at infinity
 
By the way, I stress that I mean $\mathrm{det}(R^{a}{}_b)$ not $\mathrm{det}(R_{ab})$
 
@DIRAC1930 i think the interaction potential dies off in some sense in QFT too because the particles are far apart
 
@RyderRude the exact WKB method is still not so well understood in reality
 
@DIRAC1930 while it is true such people exist, most of the standard QFT texts are not written by string theorists, and many people study QFT without any direct intentions to go on to string theory
 
@lucabtz why the mystery
 
1:33 PM
@RyderRude i mean resumming series its not at all easy
 
@Slereah Mhh yeah, it's the fully covariant tensor there
 
usually there are a lot of numerics involved in those kinda studies
 
The determinant of an endomorphism is not a density, though
 
@lucabtz oh
@DIRAC1930 just think about a classical electron in an EM field. the lagrangian technically contains a potential that doesn't die, but in reality, it does die depending on the state
 
but its interesting to study because while in QM you have both the perturbative (in h bar) series and the non-perturbative Schroedinger equation (which can be integrated at least numerically), in QFT you are left only with the perturbative series. So studying this in QM can open paths towards non-perturbative results in QFT
 
1:35 PM
if the electron is far apart from the external field (not including the self field), it behaves free
this is clear in the classical theory
the same thing is happening in qft
the self field is accounted for in the mass, and the rest of the interaction is negligible
 
Again: The main application of relativistic QFT is particle physics, not string theory, and no matter how uncomfortable we are with the various technical issues in the foundations of QFT its experimental success is undeniable - our main goal must be to find reasons why QFT works, not why it shouldn't work
2
 
@ACuriousMind I agree, all I'm stating is that many people who claim they understand QFT are just regurgitating what they have read
 
@DIRAC1930 This is uncharitable; the infuriating (to a perfectionist) truth is that none of these foundational issues matter when it comes down to computing jets in a collider or anomalous magnetic moments of particles
and again, the purpose of the standard texts is precisely to get you to be able to perform such experimentally relevant computations, not be an investigation of the foundational issues in QFT
yes, I agree with you that many people who do such computations don't "understand" QFT in the sense that I would consider "understanding"; but the fact of the matter is that such understanding seems largely superfluous for much of pragmatic physics
 
many people would also justify the computations using lattice theory
but that also has its separate problems
 
and for those of us who seek such an understanding, it is important to first understand those pragmatic applications, lest we construct beautiful theories that in the end fail to reproduce those important applications
 
1:44 PM
Really, does anyone understand QFT? What is there to understand for a human? I think all there is to accept is that is turns out we've found out the magical recipe to do calculations involving elementary particles
I think the word "understanding" itself is an overstatement
 
if u understand y those calculations work, u can generalise them
 
And that could also work with QM, I don't think there is much to understand in that sense. It's not something you derived out of logic
 
@SirCrackpot the challenge is explaining why that recipe works; if physicists just postulated the rules for Feynman diagrams etc. and went to work, that would be frustrating but unconcerning from a rigorous viewpoint - but they don't, there's all these derivations from this underlying idea of fields but none of them work fully rigorously;
 
i was about to make the Feynman diagrams comment lol
like ACuriousMind said, physics is not just calculation recipes
 
@ACuriousMind what does "unipotent gauge transformation" mean?
 
1:48 PM
physics is about finding the underlying postulates that can apply to generalised situations
 
@SirCrackpot in contrast, QM is "fully understood" on a technical level - we can haggle over interpretations for eternity, but there is no computation in non-relativistic QM that wouldn't have a fully rigorous explanation
@lucabtz I don't think this is specific to "gauge transformations", unipotent elements of a ring are the elements $A$ such that $A-I$ is nilpotent
 
@ACuriousMind yeah i read that on wiki, but what does it mean exactly for gauge transformations?
oh nevermind
i understood
 
@ACuriousMind I mean, that's where QFT stands for me right now. After you have the perturbation expansion, what comes later is merely an embellishment. Beware that I speak as someone who would like to understand the technical issues as much as you'd expect, but I think there is a dangerous line between that and understanding why nature behaves like that.
Nature just does, there is nothing to understand about that. What I mean is that either you formulate things without inconsistencies or you don't, you only have a nicer theory on a mathematical ground but ultimately nature behaves according to that very same perturbation theory that you could have postulated
 
@SirCrackpot the problem already starts with that - the perturbation expansion is asymptotic! it ultimately diverges and hence you cannot really take it as the definition of a consistent theory
if it was an ordinary convergent series, you could escape all the problems by just taking it as given, sure
 
also the perturbative expansion does not capture the non-perturbative phenomena
modulo possible resummations as i was speaking about earlier
 
1:56 PM
my point is that the kind of "understanding" I'm looking for is not an answer to the silly recursive "but why?" but to "how?", where valid answers should be both consistent and rigorous
 
@ACuriousMind Here you mean that the mathematical framework of QM is established, then we agree but I think that is not what most people are concerned about when they claim they do not understand QFT. Often it is about something that you could ask for any theory, that is: "why does nature behave like that?", they only ask for QFT because it's less intuitive
I would like to make clear that I agree about the problem of the approach to mathematical inconsistencies
 
@lucabtz that's a funny issue, actually - we always say that, but then e.g. there's famously a triangle diagram at low loop order that already computes the entire non-perturbative axial anomaly
 
these " but why" questions come under philosophy and even there, there arent any good answers
 
That's none of my issues as a physicist
 
@SirCrackpot perhaps not what most people mean, but the thing that DIRAC1930 is complaining about is that none of the standard constructions really work in deriving the existence of asymptotic states to their satisfaction; to me this is very clearly about the kind of consistency and rigor I'm talking about and not about some more nebulous "understanding"
 
1:59 PM
but @DIRAC1930 is talking about understanding why the computations work
 
@ACuriousMind yeah right, maybe should have said it does not capture all the non-perturbative phenomena
even though i never read more on why the one-loop calculation already gives the exact result for the axial anomaly
 
or rather, DIRAC is talking about understanding "how" the computations work
 
@lucabtz what I meant to say is that because we lack a proper non-perturbative formulation of QFT we can't really say what the perturbation series really "captures" with any kind of confidence
 
@ACuriousMind i agree
the axial anomaly case, if im not mistaken, is just noticed because one can compute it non-perturbatively with the Fujikawa method and ends up finding out the result from the 1-loop calculation is exactly the same
but of course its difficult to say for cases where the non-perturbative calculation is not possible
 
@lucabtz there's an extremely suspicious "accident" that there are higher "polygonal diagrams" (you encounter them if you look at anomalies in arbitrary dimensions) which cancel out for 4d theories, so only the triangle diagram remains
ever since I've seen those computations I've felt there must be something hidden in this but I don't know what
 
2:09 PM
@ACuriousMind yeah seems interesting
 
There simply sometimes are happy accidents, where lowest order corrections end up having the full contribution and higher order corrections at most just change phase. I'm not sure it is a good idea to focus on them. Instead, I think that QFT is probably on the cusp of becoming rigorously formulated. Sooner or later someone will find out how to make it all work out, i.e. no IR nor UV divergence, maybe at most with the series still being asymptotic
 
as you can tell by me not launching into a long explanation, I've not focused on it, I've simply noted it as an oddity and moved on ;)
 
Mad
i thought the potential energy of a spring is kx^2 /2, wth is kx^2 /2 -mgx , why we substracting the pot energy of the blop of mass? i am watching some MIT lecture where he does this
 
 
1 hour later…
3:26 PM
 
3:42 PM
@Mad your spring is in constant gravitational field, so there is also that contribution to the potential energy
@DIRAC1930 not only that but it's also very useful in applications. Many detectors are built to detect Cherenkov radiation
Note that I replied because you mentioned the "pot energy", which summoned me
 
Is it likely that all the relativistic effects like Cherenkov radiation have been discovered (that are available at a realistically low enough energy)
 
@Mad The lecture did not even have g.
@SirCrackpot lol
 
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